CuCl2(aq) + Na2SO3(aq) + H2O(l) → CuCl(s) + Na2SO4(aq) + 2HCl(aq). How many grams of CuCl2 and of Na2CO3 should be used if the method gives an 85.0% yield, CuCl2 is to be the limiting reactant, and a 50.0% excess of Na2SO3 is to be used?
Grams of CuCl2 =112.6g
Comments
Leave a comment